Newtonian Mechanics Review | Physics

sources:

• text: “Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). Wiley.”
• text: “Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage Learning.”

Kinematics is the study of motion without asking why it happens. Position, velocity, and acceleration are three successive derivatives of time, but they tell three different stories about the same motion. Position answers “where is the object?” — it is the raw record of location at each instant. Velocity answers “how is the position changing?” — it is the rate and direction of that change. Acceleration answers “how is the velocity changing?” — it captures the tendency of the motion itself to evolve.

The physical metaphor is a car journey. The odometer reading at each moment is the position. The speedometer reading is the velocity — how fast the position is changing. The accelerometer reading is the acceleration — whether you are pressing the brake or the accelerator, and by how much. A car moving at constant velocity has zero acceleration: the story of its motion is complete and unchanging. A car in free fall has constant acceleration g: its velocity story changes at a steady rate, and its position story is a parabola. The key insight is that acceleration is the most fundamental of the three because it is directly caused by forces (Newton’s second law), while velocity and position are consequences that follow by integration.

Newton’s second law, F=ma, is the universe’s accounting rule for motion. It states that the net force on an object equals its mass times its acceleration — or more precisely, the net force equals the rate of change of momentum. The metaphor is a financial ledger: force is the “income” or “expense” applied to an object, mass is the “inertia” (resistance to change in its state of motion), and acceleration is the resulting “change in balance.” Just as a large bank account responds slowly to small deposits, a massive object responds slowly to small forces.

The first law is the special case of zero net force: the ledger is balanced, and the object’s state of motion does not change. The third law is the double-entry bookkeeping principle: every force has an equal and opposite counterpart, so the total “debt” in the universe is always zero. The conservation laws — of momentum, energy, and angular momentum — are the consequences of this accounting system applied to isolated systems. They are the reason we can predict the outcome of a collision without knowing every microscopic detail: the ledger must balance, regardless of the complexity of the transactions.

1. First Law (Inertia): A body remains at rest or in uniform motion unless acted upon by a net force.
2. Second Law: $\mathbf{F} = m\mathbf{a} = \dot{\mathbf{p}}$ where $\mathbf{p} = m\mathbf{v}$.
3. Third Law: For every action, there is an equal and opposite reaction.

In Cartesian coordinates the component equations are straightforward:

$F_x = m\ddot{x}, \quad F_y = m\ddot{y}, \quad F_z = m\ddot{z}$

In planar polar coordinates $(r, \phi)$The acceleration decomposes into radial and transverse components:

$\mathbf{a} = (\ddot{r} - r\dot{\phi}^2)\,\hat{\mathbf{r}} + (r\ddot{\phi} + 2\dot{r}\dot{\phi})\,\hat{\boldsymbol{\phi}}$

So Newton’s second law becomes:

$F_r = m(\ddot{r} - r\dot{\phi}^2), \quad F_\phi = m(r\ddot{\phi} + 2\dot{r}\dot{\phi})$

The term $-mr\dot{\phi}^2$ is the centrifugal acceleration and $2m\dot{r}\dot{\phi}$ is the Coriolis acceleration.

In cylindrical coordinates $(\rho, \phi, z)$:

$\mathbf{a} = (\ddot{\rho} - \rho\dot{\phi}^2)\,\hat{\boldsymbol{\rho}} + (\rho\ddot{\phi} + 2\dot{\rho}\dot{\phi})\,\hat{\boldsymbol{\phi}} + \ddot{z}\,\hat{\mathbf{z}}$

Problem. A block of mass $m$ slides down an inclined plane at angle $\alpha$ to the horizontal. The coefficient of kinetic friction is $\mu_k$. Find the acceleration.

Solution. Choose axes parallel and perpendicular to the incline. The normal force is $N = mg\cos\alpha$. The friction force is $f = \mu_k N = \mu_k mg\cos\alpha$ directed up the plane. Newton’s second law along the plane:

$ma = mg\sin\alpha - \mu_k mg\cos\alpha$

$a = g(\sin\alpha - \mu_k \cos\alpha)$

The block accelerates when $\tan\alpha \gt \mu_k$ and decelerates otherwise. $\blacksquare$

Problem. A mass $m$ is attached to a string of length $l$ and rotates in a horizontal circle of radius $r$ with the string making angle $\theta$ with the vertical. Find the angular velocity $\omega$.

Solution. The forces on the mass are tension $\mathbf{T}$ along the string and weight $mg$ downward. Newton’s second law in the vertical direction:

$T\cos\theta - mg = 0 \implies T = \frac{mg}{\cos\theta}$

In the radial (horizontal) direction:

$T\sin\theta = m\omega^2 r = m\omega^2 l\sin\theta$

$\frac{mg}{\cos\theta}\sin\theta = m\omega^2 l\sin\theta$

$\omega^2 = \frac{g}{l\cos\theta}$

The period is $T = 2\pi/\omega = 2\pi\sqrt{l\cos\theta/g}$. $\blacksquare$

Theorem 1.1 (Conservation of Linear Momentum). For a system of $N$ particles with no external forces, the total linear momentum is conserved.

Proof. Newton’s second law for the $i$-th particle:

$\mathbf{F}_i^{(\mathrm{ext})} + \sum_{j \neq i} \mathbf{F}_{ij} = m_i \dot{\mathbf{v}}_i$

Where $\mathbf{F}_{ij}$ is the force on particle $i$ due to particle $j$. By Newton’s third law, $\mathbf{F}_{ij} = -\mathbf{F}_{ji}$. Summing over all particles:

$\sum_i \mathbf{F}_i^{(\mathrm{ext})} + \sum_i \sum_{j \neq i} \mathbf{F}_{ij} = \frac{d}{dt}\sum_i m_i \mathbf{v}_i$

The double sum vanishes by Newton’s third law. Defining $\mathbf{P} = \sum_i m_i \mathbf{v}_i$:

$\sum_i \mathbf{F}_i^{(\mathrm{ext})} = \dot{\mathbf{P}}$

If there are no external forces, $\dot{\mathbf{P}} = 0$ and $\mathbf{P}$ is constant. $\blacksquare$

Corollary. The centre of mass moves as if all external forces acted on a single particle of mass $M = \sum_i m_i$ located at the centre of mass: $M\ddot{\mathbf{R}} = \sum_i \mathbf{F}_i^{(\mathrm{ext})}$.

Theorem 1.2 (Work-Energy Theorem). The work done by the net force on a particle equals the change in its kinetic energy:

$W = \int_{\mathbf{r}_1}^{\mathbf{r}_2} \mathbf{F} \cdot d\mathbf{r} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

Proof. Using Newton’s second law:

$W = \int m\mathbf{a} \cdot \mathbf{v}\, dt = \int m \frac{d\mathbf{v}}{dt} \cdot \mathbf{v}\, dt = \int m\mathbf{v} \cdot d\mathbf{v} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

$\blacksquare$

Definition. A force is conservative if the work done is path-independent, equivalently $\nabla \times \mathbf{F} = \mathbf{0}$Equivalently $\mathbf{F} = -\nabla V$ for some scalar potential $V(\mathbf{r})$.

Theorem 1.3 (Conservation of Mechanical Energy). If all forces are conservative, $E = T + V$ is conserved.

Proof. For a conservative force, $W = -\Delta V$. By the work-energy theorem:

$-\Delta V = \Delta T \implies \Delta(T + V) = 0$

$\blacksquare$

Theorem 1.4 (Conservation of Angular Momentum). If the net external torque on a system vanishes, the total angular momentum is conserved.

Proof. The angular momentum of the $i$-th particle about the origin is $\mathbf{L}_i = \mathbf{r}_i \times m_i \mathbf{v}_i$. Taking the time derivative:

$\dot{\mathbf{L}}_i = \dot{\mathbf{r}}_i \times m_i \mathbf{v}_i + \mathbf{r}_i \times m_i \dot{\mathbf{v}}_i = \mathbf{r}_i \times \mathbf{F}_i$

Since $\dot{\mathbf{r}}_i \times m_i \mathbf{v}_i = \mathbf{v}_i \times m_i \mathbf{v}_i = \mathbf{0}$. Summing over all particles:

$\dot{\mathbf{L}} = \sum_i \mathbf{r}_i \times \mathbf{F}_i^{(\mathrm{ext})} + \sum_i \sum_{j \neq i} \mathbf{r}_i \times \mathbf{F}_{ij}$

The double sum represents internal torques. For central internal forces ($\mathbf{F}_{ij}$ parallel to $\mathbf{r}_i - \mathbf{r}_j$), the internal torques cancel in pairs. Hence:

$\dot{\mathbf{L}} = \boldsymbol{\tau}^{(\mathrm{ext})}$

If $\boldsymbol{\tau}^{(\mathrm{ext})} = \mathbf{0}$ Then $\mathbf{L} = \mathrm{const}$. $\blacksquare$

Definition. The rocket equation (Tsiolkovsky equation) describes the motion of a rocket that expels mass at a constant exhaust velocity.

Consider a rocket of mass $m$ moving with velocity $v$ in one dimension. In time $dt$It ejects mass $dm$ (where $dm \lt 0$) at exhaust velocity $u_e$ relative to the rocket. The ejected mass has velocity $v - u_e$ in the lab frame. By conservation of momentum:

$mv = (m + dm)(v + dv) + (-dm)(v - u_e)$

Neglecting the second-order term $dm\, dv$:

$mv = mv + m\, dv + dm\, v - dm\, v + u_e\, dm$

$0 = m\, dv + u_e\, dm$

$dv = -u_e \frac{dm}{m}$

Integrating from initial mass $m_0$ and velocity $v_0$ to final mass $m_f$ and velocity $v_f$:

$v_f - v_0 = u_e \ln\frac{m_0}{m_f}$

This is the Tsiolkovsky rocket equation.

Theorem 1.5 (Rocket Equation with Gravity). If the rocket moves vertically against a uniform gravitational field $g$:

$\Delta v = u_e \ln\frac{m_0}{m_f} - g\, \Delta t$

Where $\Delta t$ is the burn time.

Problem. A rocket starts from rest with mass $m_0 = 1000\,\mathrm{kg}$ and exhaust velocity $u_e = 3000\,\mathrm{m}/s$. It burns fuel until its mass is $m_f = 400\,\mathrm{kg}$. Find the final velocity.

Problem. A particle of mass $m_1 = 2\,\mathrm{kg}$ moving at $\mathbf{v}_1 = 4\hat{\mathbf{i}}\,\mathrm{m/s}$ collides elastically with a particle of mass $m_2 = 3\,\mathrm{kg}$ at rest. After the collision, $m_1$ moves at $30^\circ$ to the $x$-axis. Find the final velocities.

Common mistake. Assuming the second particle moves along the $x$-axis. In two-dimensional elastic collisions, both particles generally move at angles to the original direction.

Problem. A block of mass $m = 5\,\mathrm{kg}$ sits on a frictionless horizontal surface inside an elevator accelerating upward at $a = 2\,\mathrm{m/s^2}$. A horizontal force $F = 20\,\mathrm{N}$ is applied. Find the acceleration relative to the elevator.

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